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About two integrals in QCD textbook by muta 
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#1
May714, 11:25 PM

P: 9

1.How to deal with the delta functions in eq.2.3.153 to obtain the eq.2.3.154 by integrating over q'? 2.How to caculate the integral from eq.2.3.154 to eq.2.3.156, especially the theta function? 


#2
May814, 09:48 PM

P: 305

You already have a [itex]\delta^4(q'q)[/itex] so just put q=q' and remove the integral over dq'. Then in Eq. (2.3.154) you have [itex]\delta(k'\cdot q)[/itex] that, in the COM, becomes just [itex]\delta(k_0\sqrt{s})[/itex]. Therefore, taking care of the Jacobian coming out of the delta function, this just tells you that [itex]k'_0=0[/itex]. Hence you don't need to worry about the thetas since they only give you [itex]\theta(q_0)=\theta(\sqrt{s})=1[/itex] since its argument is positive.



#3
May914, 04:06 AM

P: 9

Actually, when I try to simplify the eq.2.3.153 for obtain the bracket in eq.2.3.154, I find we must use the two delta functions in eq.2.3.154, but I wonder whether we can use δ(1/4(q'+k')^2m^2)=δ(k'^2+s4m^2)δ(k'*q) directly, and the δ(k*q) is equivalent the δ(k'*q) because of the k' is integrated over total space?



#4
May914, 09:00 AM

P: 305

About two integrals in QCD textbook by muta
$$ \delta\left(q^2+k'^2+2q\cdot k'4m^2\right)\delta\left(q^2+k'^22q\cdot k'4m^2\right). $$ Now keep in mind that if you have some function [itex]f(x)[/itex] multiplied by a delta then [itex]\delta(xx_0)f(x)=\delta(xx_0)f(x_0)[/itex]. This is true also if the function is a delta itself. Now, the second delta tells you that [itex]q^2+k'^24m^2=2q\cdot k'[/itex]. Using this is the first delta you obtain (I always omit the necessary Jacobian): $$ \delta\left(q\cdot k'\right)\delta\left(q^2+k'^22q\cdot k'4m^2\right)=\delta\left(q\cdot k'\right)\delta\left(q^2+k'^24m^2\right). $$ Keep also in mind that [itex]q^2=s[/itex]. 


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